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Question

If |z1+z2|=|z1z2| then argz1argz2=

A
(2n+1)π2,nZ
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B
nπ+π4,nZ
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C
2nπ,nZ
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D
(2n+1)π,nZ
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Solution

The correct option is A (2n+1)π2,nZ
Given that |z1+z2|=|z1z2|
|z1+z2|2=|z1z2|2
[|z1z2|2=|z1|2+|z2|22Re(z1¯¯¯¯¯z2 ),|z1+z2|2=|z1|2+|z2|2+2Re(z1¯¯¯¯¯z2 )]
|z1|2+|z2|2+2Re(z1¯¯¯¯¯z2 )=|z1|2+|z2|22Re(z1¯¯¯¯¯z2 )
4Re(z1¯¯¯¯¯z2 )=0
Re(z1¯¯¯¯¯z2 )=0

Let z1=r1eiθ1 and z2=r2eiθ2
Then z1¯¯¯¯¯z2=r1r2eiθ1eiθ2=r1r2ei(θ1θ2)
z1¯¯¯¯¯z2=r1r2(cos(θ1θ2)+isin(θ1θ2))
Re(z1¯¯¯¯¯z2 )=0
cos(θ1θ2)=0
θ1θ2=(2n+1)π2,nZ

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